What is the integral of 1+sin^2(x) ?

The integral of 1+sin^2(x) is 3/2 x-1/4 sin(2x)+C

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integral of 1+sin^2(x)

\int 1 + sin^{2} (x) d x

\frac{3}{2} x - \frac{1}{4} sin (2 x) + C

Solution steps

\int 1 + sin^{2} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d x + \int sin^{2} (x) d x

\int 1 d x = x

\int sin^{2} (x) d x = \frac{1}{2} (x - \frac{1}{2} sin (2 x))

= x + \frac{1}{2} (x - \frac{1}{2} sin (2 x))

Simplify

x + \frac{1}{2} (x - \frac{1}{2} sin (2 x)) : \frac{3}{2} x - \frac{1}{4} sin (2 x)

= \frac{3}{2} x - \frac{1}{4} sin (2 x)

Add a constant to the solution

= \frac{3}{2} x - \frac{1}{4} sin (2 x) + C

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What is the integral of 1+sin^2(x) ?
The integral of 1+sin^2(x) is 3/2 x-1/4 sin(2x)+C